Polar Coordinate function of a Straight Line

$x = r\cos \theta\\ y = r\sin\theta\\ 2x + 3y - 6 = 0\\ 2r\cos\theta + 3r\sin\theta = 6\\ r(2\cos\theta + 3\sin\theta) = 6$

Now I could say:

$r = \frac {6}{3\cos\theta + 2\sin\theta}$

and be done.

But I think that this is a little bit more informative:

$\sin (\arctan \frac ab) = \frac {a}{\sqrt {a^2 + b^2}}\\ \cos (\arctan \frac ab) = \frac {b}{\sqrt {a^2 + b^2}}$

$r\sqrt {2^2 + 3^2}(\sin (\arctan \frac{3}{2})\sin\theta + \cos(\arctan \frac 32)\cos\theta) = 6\\ r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\ r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\ r = \frac {6}{\sqrt {13}} \sec (\theta - \arctan \frac 32)$

As it gives the angle of rotation and the distance to the line.

In Cartesian coordinates, a straight line equation is $y=mx+b$ where is $m$ is a numerical slope and $b$ is a numerical $y$ intercept. Following rules for converting to polar coordinates, we let $x=r\cdot cos\theta$ and $y=r\cdot sin\theta$. Solve for $r$ $$\left\{ r=-\frac{b}{m\;\operatorname{cos}\left(\theta\right)-\operatorname{sin}\left(\theta\right)}\right\}$$

I typically use $0\le\theta\le 2\pi$ The idea of graphing a line in polar coordinates is more of a curiosity to me than anything practical. In Geogebra, (for example) there is not really a polar graphing system, but we can plot a polar equation by conversion to cartesian. that is, $x=r(\theta)\cos\theta$ and $y=r(\theta)\sin\theta$. Geogebra then superimposes a polar grid and viola it looks like we have graphed a polar equation. So... No real restriction on $r$, except that $\theta$ will have to have exceedingly high precision at certain points in its range or you will jump over the large values of $r(\theta)$. I would have to play with this awhile to give you any better answer.

The straight line through $(0,2)\text{ and } (3,0)$ has equation $$y=-\frac{2}{3}x+2$$ Since the the polar coordinates $(r,\alpha)$ correspond to $(r\sin\alpha,r\cos\alpha)$ this translates to $$ -\frac{2}{3}r\cos\alpha = r\sin\alpha+2$$

It only remains to solve for $r$.